Q9DQ - University Physics with Modern Physics

Question

projectile is fired upward at an angle $\mathit{\theta }$ above the horizontal with an initial speed ${\mathit{v}}_{\mathbf{0}}$ . At its maximum height, what are its velocity vector, its speed, and its acceleration vector? 

Step-by-Step Solution

Answer

At maximum height, the velocity vector is $v_0 \cos \theta \hat{i}$ , speed of projectile is $v_0 \cos \theta$ , and acceleration vector is $-g \hat{j}$.

Step 1: Identification of given data

The given data can be listed below,

• Angle of projectile is, $\theta$.
• The initial speed of the projectile is, $v_0$ .

Step 2: Concept/Significance of projectile motion

The initial launch speed, launch angle, and acceleration due to gravity are the only factors that determine projectile velocity.

Step 3: Determination of velocity vector, its speed, and its acceleration vector.

The path for the projectile is given by,

                                                     

As the vertical acceleration (acceleration due to gravity) acts in a downward direction, the vertical component of velocity steadily diminishes. When the bullet reaches the highest point on its course, the vertical component of velocity vanishes, leaving just the horizontal component. The velocity works along the horizontal, which is the positive x-axis, at maximum height. As a result, at the maximum height position, the vector representation of velocity vector is,

${\overrightarrow{v}}_H=v_x \hat{i}$                                                                                                                            …(i)

Here, $v_{0.x}$ is the horizontal component of velocity at maximum height and $\hat{i}$ is the unit vector along the x-axis. The horizontal component of launch velocity remains constant along the path because no acceleration is acting on the projectile along the horizontal direction. So, the horizontal velocity is given by,

$v_x=v_0 \cos \theta$

Substitute the values of $v_x$ into equation (i).

${\overrightarrow{v}}_H=v_0 \cos \theta \hat{i}$

The magnitude of velocity is given by,

$$\begin{gathered} \left|{\overrightarrow{v}}_H\right|=\left|v_0 \cos \theta \hat{i}\right| \\ {v}_H=v_0 \cos \theta \end{gathered}$$

The acceleration along the horizontal axis is zero so the component of the vertical acceleration is given y,

$a_y=g$

The net acceleration is given by,

${\overrightarrow{a}}_H=a_x \hat{i}+a_y\left(-\hat{j} \right)$

Here, j is the unit vector along the y-axis and the negative sign indicates that the vertical acceleration is acting in downward direction.

Substitute all the values of horizontal and vertical acceleration is given by,

$$\begin{gathered} {\overrightarrow{a}}_H=0+g\left(-\hat{j} \right) \\ =-g \hat{j} \end{gathered}$$

Thus, at maximum height, the velocity vector is $v_0 \cos \theta \hat{i}$ , speed of projectile is $v_0 \cos \theta$, and acceleration vector is $-g \hat{j}$ .